Question: Simplify and expand the following expression: $ \dfrac{1}{r - 6}- \dfrac{1}{4r + 24}- \dfrac{1}{r^2 - 36} $
Solution: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $4$ out of denominator in the second term: $ \dfrac{1}{4r + 24} = \dfrac{1}{4(r + 6)}$ We can factor the quadratic in the third term: $ \dfrac{1}{r^2 - 36} = \dfrac{1}{(r - 6)(r + 6)}$ Now we have: $ \dfrac{1}{r - 6}- \dfrac{1}{4(r + 6)}- \dfrac{1}{(r - 6)(r + 6)} $ The least common multiple of the denominators is: $ (r - 6)(r + 6)$ In order to get the first term over $(r - 6)(r + 6)$ , multiply by $\dfrac{4(r + 6)}{4(r + 6)}$ $ \dfrac{1}{r - 6} \times \dfrac{4(r + 6)}{4(r + 6)} = \dfrac{4(r + 6)}{(r - 6)(r + 6)} $ In order to get the second term over $(r - 6)(r + 6)$ , multiply by $\dfrac{r - 6}{r - 6}$ $ \dfrac{1}{4(r + 6)} \times \dfrac{r - 6}{r - 6} = \dfrac{r - 6}{(r - 6)(r + 6)} $ In order to get the third term over $(r - 6)(r + 6)$ , multiply by $\dfrac{4}{4}$ $ \dfrac{1}{(r - 6)(r + 6)} \times \dfrac{4}{4} = \dfrac{4}{(r - 6)(r + 6)} $ Now we have: $ \dfrac{4(r + 6)}{(r - 6)(r + 6)} - \dfrac{r - 6}{(r - 6)(r + 6)} - \dfrac{4}{(r - 6)(r + 6)} $ $ = \dfrac{ 4(r + 6) - (r - 6) - 4} {(r - 6)(r + 6)} $ Expand: $ = \dfrac{4r + 24 - r + 6 - 4}{4r^2 - 144} $ $ = \dfrac{3r + 26}{4r^2 - 144}$